Answer
See below
Work Step by Step
Given:
$$e^{At}=diag(e^{d_1t},e^{d_1t},...e^{d_nt}$$
According to Taylor series we have:
$$e^{At}=I_n+At+\frac{1}{2!}(At)^2...+\frac{1}{k!}(At)^k+...\\
=diag(1,...1)+diag(d_1,...d_n)+\frac{1}{2!}diag(d_1^2,...d_n^2)+...+\frac{1}{k!}diag(d^k_1,...,d^k_n)\\
=diag(1+d_1+\frac{1}{2}d^2_1+...+1+d_n+\frac{1}{2!}d^2_n+...+\frac{1}{k!}d^k_n) \\
=diag(e^{d_1t},e^{d_2t},e^{d_3t}...,e^{d_nt})$$
